# Primitive of x squared over x squared minus a squared/Logarithm Form

## Theorem

$\displaystyle \int \frac {x^2 \ \mathrm d x} {x^2 - a^2} = x + \frac a 2 \ln \left({\frac {x - a} {x + a} }\right) + C$

for $x^2 > a^2$.

## Proof

Let:

 $\displaystyle \int \frac {x^2 \ \mathrm d x} {x^2 - a^2}$ $=$ $\displaystyle \int \frac {x^2 - a^2 + a^2} {x^2 - a^2} \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \int \frac {x^2 - a^2} {x^2 - a^2} \ \mathrm d x + \int \frac {a^2} {x^2 - a^2} \ \mathrm d x$ $\displaystyle$ $=$ $\displaystyle \int \mathrm d x + a^2 \int \frac {\mathrm d x} {x^2 - a^2}$ Primitive of Constant Multiple of Function $\displaystyle$ $=$ $\displaystyle x + a^2 \int \frac {\mathrm d x} {x^2 - a^2} + C$ Primitive of Constant $\displaystyle$ $=$ $\displaystyle x + a^2 \left({\frac 1 {2 a} \ln \left({\frac {x - a} {x + a} }\right)}\right) + C$ Primitive of Reciprocal of $x^2 - a^2$: Logarithm Form $\displaystyle$ $=$ $\displaystyle x + \frac a 2 \ln \left({\frac {x - a} {x + a} }\right) + C$ simplifying

$\blacksquare$